On Mon, Dec 08, 2014 at 09:44:22AM -0800, Ryan Carboni wrote: > 24 = hours > 60 = minutes > 7 = days > > so I'm only off by a factor of 2^3.3, not by a factor of 2^9.3 > > > Cheers. >
Isn't this enough to find 128 bit md5 collision? Appears to me they can do it distributed in about 2 days even with the most naive rho attack. AFAIK it is open problem if 128 bit md5 collision exists (though it is believed to exist). > On Mon, Dec 8, 2014 at 3:23 AM, Joseph Birr-Pixton <[email protected]> > wrote: > > > On 8 December 2014 at 10:40, Ryan Carboni <[email protected]> wrote: > > > https://blockexplorer.com/q/hashestowin > > > log(171833398380382098659*24*60*7)/log(2) > > > > I think your calculation is slightly off. hashestowin is the average > > number of hashes you need to perform to win the current block. It's > > not necessarily the case that a block is calculated each second: in > > fact one is found (on average) each 625 seconds[1]. > > > > So that gives: > > > > >>> log(171833398380382098659*24*60*7/625)/log(2) > > 71.23 > > > > Cheers, > > Joe > > > > [1] https://blockexplorer.com/q/interval > >
